Lesson_5.5_Homework_Solution

# Lesson_5.5_Homework_Solution - Lesson 5.5 Homework...

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Lesson 5.5 Homework Solutions 1. The resultant amplitude is given by: 2 cos 2 0.02 cos 2 12 0.039 mm s A δ π = = × × = 2. (i) Resultant amplitude = 2 cos 2 0.05 cos 2 4 0.07 mm A = × × = (ii) Resultant amplitude = 2 cos 2 0.05 0 cos 2 2 A = × × = 3. (i) -1 340 ms 68 z . H 5 0 v f m λ = = = (ii) Path difference x = 11.2 – 10 = 1.2 m 1.2 m 5 m 0.24 x = = (iii) The phase difference that corresponds to a path difference of λ = 2 π Phase difference for 0.24 λ = 0.48 π (iv) The resultant amplitude = 2 cos 2 0.01 cos0.2 0.015 4 2 m A m = × × = 4. 40.07 40 2.4 Path difference 40.07 – 40 = 0.07 m (i) The condition for minimum sound is: (2 1) 0.07 2 n + = For this condition, the frequency will be minimum when n = 0

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Then, λ /2 = 0.07, or λ = 0.14 m 340 2429 0.14 v f Hz λ = = = The condition for maximum sound is n λ = 0.07 The minimum frequency for this condition occurs when n = 1 Then λ = 0.07 340 0.07 4857 v Hz f = = = 5. 442 Hz or 438 Hz
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Lesson_5.5_Homework_Solution - Lesson 5.5 Homework...

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